Question

The resultant force on the square current loop PQRS due to a long current-carrying conductor will be:

Answer 1

Step 1: Given that

The following arrangement is given in the diagram below.

$$\begin{gathered} I_1=30A \\ {I}_2=20A \\ {\mu }_0=4\pi \times {10}^{-7} \\ L=10\times {10}^{-2} \end{gathered}$$

Here,

$I$ is the current.

$L$ is the length.

Since the current in the lengths, PQ and RS are equal and opposite, the forces also will be equal and opposite. Hence both will cancel each other.

Step 2: Formula used

$F=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}$

Where,

$d$ is the distance between the two parallel conductors.

Step 3: Calculating the force on the loop

The force on wire PQ and SR will cancel each other since the current flowing in the identical wires are the same.

The force on PS is:

$F_{PS}=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}=\frac{4\pi \times {10}^{-7}\times 30\times 20\times 10\times {10}^{-2}}{2\pi \times 2\times {10}^{-2}}$

$F_{PS}=-6\times {10}^{-4}\overrightarrow{i}$

The force on QR is:

$F_{QR}=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}=\frac{4\pi \times {10}^{-7}\times 30\times 20\times 10\times {10}^{-2}}{2\pi \times 12\times {10}^{-2}}$

$F_{QR}=1\times {10}^{-4}\overrightarrow{i}$

Hence the net force is

$$\begin{gathered} F_{net}=F_{PS}+F_{QR} \\ =\left(-6\times {10}^{-4}\overrightarrow{i}\right)+1\times {10}^{-4}\overrightarrow{i} \\ =-5\times {10}^{-4}N\overrightarrow{i} \end{gathered}$$

Hence, the resultant force on the square current loop PQRS due to a long current-carrying conductor will be $-5\times {10}^{-4}N\overrightarrow{i}$.